Answer: Option A: The number of trees sampled.
The accuracy can be understood as how close is the measured value to the true value. The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.
Answer:
the pe at the top of the building: 784 J
the pe halfway through the fall: 392 J
the pe just before hitting the ground: 784 J
Explanation:
Pls brainliest me
I had this question before
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration 
directed downward, initial vertical position 
, and initial vertical velocity 
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
substituting, we find
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
In order to do that, we use again the same SUVAT equation substituting 
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
Which means that the heigth of the packet was
R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4
R2 + R5 = 1350 + 1150 = 2500 = R25
The circuit has been reduced to 3 resistors in parallel
R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3
R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler
Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So 
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
